2
$\begingroup$

Evaluation of $$\lim_{x\rightarrow \frac{\pi}{2}}\frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\ln (\sin x)}$$

Without Using L hopital Rule and series expansion.

$\bf{My\; Try::}$ I have solved it using L hopital Rule and series expansion.

But I did not undersand How can i solve it Without Using L hopital Rule and series expansion

Help required, Thanks

$\endgroup$
  • $\begingroup$ You increase the difficulty of the problem many fold by restricting the tools to be used. +1 $\endgroup$ – Paramanand Singh Jun 22 '16 at 9:27
1
$\begingroup$

Most of the common limit problems can be solved without the use of L'Hospital's Rule and Taylor's series (i.e. without using differentiation in anyway). This limit is not one of them. I have avoided differentiation but at the cost of using the following limit: $$\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}} = -\frac{1}{2}\tag{1}$$ apart from the following standard limits $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}\tag{2}$$ The limit $(1)$ is established at the end of this answer using a specific definition of $e^{x}$.


Let's put $\sin x = t$ so that as $x \to \pi/2$ we have $t \to 1$. Now the desired limit is evaluated as follows \begin{align} L &= \lim_{x \to \pi/2}\frac{\sin x - (\sin x)^{\sin x}}{1 - \sin x + \log(\sin x)}\notag\\ &= \lim_{t \to 1}\frac{t - t^{t}}{1 - t + \log t}\notag\\ &= \lim_{t \to 1}\frac{t - \exp(t\log t)}{1 - t + \log t}\notag\\ &= \lim_{h \to 0}\frac{1 + h - \exp((1 + h)\log(1 + h))}{\log(1 + h) - h}\text{ (putting }t = 1 + h)\notag\\ &= \lim_{h \to 0}\dfrac{1 + h - \exp((1 + h)\log(1 + h))}{\dfrac{\log(1 + h) - h}{h^{2}}\cdot h^{2}}\notag\\ &= -2\lim_{h \to 0}\dfrac{1 + h - \exp(\log(1 + h) + h\log(1 + h))}{h^{2}}\text{ (using (1))}\notag\\ &= -2\lim_{h \to 0}\dfrac{1 + h - (1 + h)\cdot\exp(h\log(1 + h))}{h^{2}}\notag\\ &= -2\lim_{h \to 0}\dfrac{1 + h - (1 + h)\cdot\exp(h\log(1 + h))}{h^{2}}\notag\\ &= 2\lim_{h \to 0}(1 + h)\cdot\dfrac{\exp(h\log(1 + h)) - 1}{h^{2}}\notag\\ &= 2\lim_{h \to 0}\dfrac{\exp(h\log(1 + h)) - 1}{h\log(1 + h)}\cdot\frac{\log(1 + h)}{h}\notag\\ &= 2 \cdot 1 \cdot 1 = 2\notag \end{align}


To establish $(1)$ we first establish $$\lim_{x \to 0}\frac{e^{x} - 1 - x}{x^{2}} = \frac{1}{2}\tag{3}$$ via definition $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{4}$$ and then we have \begin{align} A &= \lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\notag\\ &= \lim_{t \to 0}\dfrac{t - e^{t} + 1}{(e^{t} - 1)^{2}}\text{ (putting }t = \log(1 + x))\notag\\ &= \lim_{t \to 0}\dfrac{t - e^{t} - 1}{\left(\dfrac{e^{t} - 1}{t}\right)^{2}\cdot t^{2}}\notag\\ &= \lim_{t \to 0}\dfrac{t - e^{t} + 1}{t^{2}}\notag\\ &= -\frac{1}{2}\text{ (using (3))}\notag \end{align}


To prove $(3)$ note that using $(4)$ and binomial theorem for positive integral index we have $$e^{x} = \lim_{n \to \infty}1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x^{3} + \cdots$$ and hence $$\frac{e^{x} - 1 - x}{x^{2}} = \lim_{n \to \infty}\dfrac{1 - \dfrac{1}{n}}{2!} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}x + \cdots = \lim_{n \to \infty}f(x, n)\text{ (say)}\tag{5}$$ where $f(x, n)$ is a finite sum. If $0 < x < 1$ then $$\frac{n - 1}{2n} \leq f(x, n) \leq \frac{1}{2} + \frac{x}{6} + \frac{x^{2}}{24} + \cdots + \frac{x^{n - 2}}{n!}$$ and hence $$\frac{n - 1}{2n} \leq f(x, n) \leq \frac{1}{2} + \frac{x}{6} + \frac{x^{2}}{18} + \cdots = \frac{3}{6 - 2x}$$ Taking limit of the above inequality as $n \to \infty$ we get using $(5)$ $$\frac{1}{2} \leq \frac{e^{x} - 1 - x}{x^{2}} \leq \frac{3}{6 - 2x}$$ for $0 < x < 1$. Now taking limit as $x \to 0^{+}$ we get $$\lim_{x \to 0^{+}}\frac{e^{x} - 1 - x}{x^{2}} = \frac{1}{2}$$ and the case for $x \to 0^{-}$ is handled by putting $x = -y$.


The above long proofs of limits $(1)$ and $(3)$ show that Taylor's series and L'Hospital's Rule are very powerful tools which can replace these long proofs to just one step evaluation. But there is an intrinsic interest in establishing limits without the use of differentiation and I hope the above presentation is a good attempt in that direction.

$\endgroup$
0
$\begingroup$

It is not a direct answer. By substition $x-\frac { \pi }{ 2 } =t$ $$f\left( x \right) =\frac { \cos { x } -{ \cos { x } }^{ \cos { x } } }{ 1-\cos { x } +\ln { \cos { x } } } $$

$$\lim _{ t\rightarrow 0 }{ \frac { \sin { \left( \frac { \pi }{ 2 } +t \right) - } { \sin { \left( \frac { \pi }{ 2 } +t \right) } }^{ \sin { \left( \frac { \pi }{ 2 } +t \right) } } }{ 1-\sin { \left( \frac { \pi }{ 2 } +t \right) +\ln { \left( \sin { \left( \frac { \pi }{ 2 } +t \right) } \right) } } } } =\lim _{ t\rightarrow 0 }{ \frac { \cos { t } -{ \cos { t } }^{ \cos { t } } }{ 1-\cos { t } +\ln { \cos { t } } } } $$ enter image description here

Analysing graph of the function i think (i hope) we can get an answer

$\endgroup$
  • 1
    $\begingroup$ But then you could also just plot the original function around $x=\pi/2$...? $\endgroup$ – StackTD Jun 21 '16 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.